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From a point $O$ on the circle $x^2+y^2=d^2$, tangents $OP$ and $OQ$ are drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b$. Show that the locus of the midpoint of chord PQ is given by $$x^2+y^2=d^2\bigg[\frac{x^2}{a^2}+\frac{y^2}{b^2}\bigg]^2$$

I recognize that the locus of a chord whose midpoint is at $(h,k)$ is given by $\frac{xh}{a^2}+\frac{yk}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$

I also recognize that PQ is the chord of contact, but to find its equation using the chord of contact formula I would require the coordinates of point O which I do not have.

Here I am getting the equation in terms of $x,y,h,k$, but to find the locus I need the equation entirely in the form of $h,k$, right? So how do I eliminate $x,y$ from the equation of the locus of the midpoint?

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This may not be the solution you are looking for:

Let $\mathcal C=\{x^2+y^2=1\}$ be the unit circle. Let $O' = (\alpha, \beta)$ be any point outside of this circle. Let $O'P'$ and $O'Q'$ be two tangent line to $\mathcal C$. One can check that the midpoint $m' = (x, y)$ of $P'Q'$ is given by (why?)

$$(x, y) = m' = \frac{1}{\alpha^2+ \beta^2} (\alpha, \beta).$$

Now assume that $O = (\alpha, \beta)$ is on the ellipse $\{ (ax)^2 + (by)^2 = d^2\}$. Thus $(a\alpha)^2 + (b\beta)^2 = d^2$. Then

$$ d^2(x^2 + y^2)^2 = \frac{d^2}{(\alpha^2 + \beta^2)^2}$$

and

$$(ax)^2 +(by)^2 = \frac{(a\alpha)^2 + (by)^2}{(\alpha^2 + \beta^2)^2}=\frac{d^2}{(\alpha^2 + \beta^2)^2}$$

Thus the locus of the midpoint $m'$ is given by

$$ \tag{1} (ax)^2 + (by)^2 = d^2 (x^2+ y^2)^2. $$

The above is related to your question in the following way: Consider the transformation: $$(x, y) \mapsto (x/a, y/b).$$ Under this transformation, the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ is sent to the unit circle $\mathcal C$, while the circle $x^2 + y^2 = d^2$ is sent to the ellipse $(ax)^2 + (by)^2 = d^2$. The crucial observation is that tangent lines $OP, OQ$ are also sent to tangent lines $O'P', O'Q'$, and the midpoint $m$ of $PQ$ are sent to the midpoint $m'$ of $P'Q'$ (see here). Thus if you take the inverse transformation

$$ (x, y) \mapsto (ax, by)$$

Then the locus of $m'$ will be sent to the locus of $m$. This implies your equation: if you change $x$, $y$ to $x/a$, $y/b$ respective in (1), you get

$$x^2 + y^2 = d^2 \left(\frac{x^2}{a^2}+ \frac{y^2}{b^2}\right)^2.$$

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  • $\begingroup$ Thanks but I was looking for something starting with what I used $\endgroup$
    – Techie5879
    Nov 8 '19 at 8:56
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Still not what you want, but uses Joachimsthals notations.

The locus is the midpoint of the two intersection points of $$s=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$$ and (from $s_1^2=s \cdot s_{11}$) $$(\frac{x(O)x}{a^2}+\frac{y(O)y}{b^2}-1)^2=(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1)(\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}-1)$$ which (since the square roots cancel in $\frac{x_1+x_2}{2}$ and $\frac{y_1+y_2}{2}$) is $$(x,y)=(\frac{x(O)}{\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}},\frac{y(O)}{\frac{x(O)^2}{a^2}+\frac{y(O)^2}{b^2}}),$$ where $x(O)^2+y(O)^2=d^2$ since $O$ is on that circle.

Writing $h=x(O), k=y(O)$ into M2

R=QQ[a,b,d]
S=R[h,k,x,y,MonomialOrder=>Eliminate 2]
I=ideal(h^2+k^2-d^2,(b^2*h^2+a^2*k^2)*x-a^2*b^2*h,(b^2*h^2+a^2*k^2)*y-a^2*b^2*k)
gens gb I

yields $$a^6b^6d^2(d^2(\frac{x^2}{a^2}+\frac{y^2}{b^2})^2-(x^2+y^2)).$$

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